An element 'X' (At. Mass = 40 g `mol^(-1)`) having f.c.c structure has unit cell edge length of 400 pm . Calculate the density of 'X' and the number of unit cells in 4 g of 'X' .
`(N_(A) = 6.022 xx 10^(23) mol^(-1))`.

Answer

`4.15 g cm^(-3), 1.505 xx 10^(22)`

Explanation

`d = (Z xx M)/(a^3 xx N_A)`
`= (4 xx 40)/((400 xx 10^(-10))^3 xx 6.022 xx 10(23))`
`= 4.15 g cm^(-3)`
40 g of element = `6.022 xx 10^(23)` atoms
4 g of element = `6.022 xx 10^(23)` atoms
`= 6.022 xx 10^(22)` atoms
For fcc, Z = 4
1 unit cell = 4 atoms
No. of unit cells in `4g = (6.022 xx 10^(22))/(4)`
`= 1.505 xx 10^(22)`